Question: Complete the square to solve for $x$. $x^{2}+18x+80 = 0$
Explanation: Begin by moving the constant term to the right side of the equation. $x^2 + 18x = -80$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $18$ , half of it would be $9$ , and squaring it gives us ${81}$ $x^2 + 18x { + 81} = -80 { + 81}$ We can now rewrite the left side of the equation as a squared term. $( x + 9 )^2 = 1$ Take the square root of both sides. $x + 9 = \pm1$ Isolate $x$ to find the solution(s). $x = -9\pm1$ So the solutions are: $x = -8 \text{ or } x = -10$ We already found the completed square: $( x + 9 )^2 = 1$